\newproblem{lay:7_2_1}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.2.1}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Compute the quadratic form $\mathbf{x}^TA\mathbf{x}$, when $A=\begin{pmatrix} 5 & \frac{1}{3} \\ \frac{1}{3} & 1\end{pmatrix}$ and
	\begin{enumerate}[a.]
		\item $\mathbf{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}$
		\item $\mathbf{x}=\begin{pmatrix}6\\1\end{pmatrix}$
		\item $\mathbf{x}=\begin{pmatrix}1\\3\end{pmatrix}$
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item We simply need to perform all the multiplications
		      \begin{center}
						$\begin{array}{rcl}Q(\mathbf{x})&=&\mathbf{x}^TA\mathbf{x}\\
						   &=&\begin{pmatrix}x_1 &x_2\end{pmatrix}\begin{pmatrix} 5 & \frac{1}{3} \\ \frac{1}{3} & 1\end{pmatrix}
						       \begin{pmatrix}x_1\\x_2\end{pmatrix} \\
						   &=&\begin{pmatrix}x_1 &x_2\end{pmatrix}\begin{pmatrix}5x_1+\frac{1}{3}x_2 \\ \frac{1}{3}x_1+x_2\end{pmatrix}\\
							 &=&5x_1^2 + \frac{2}{3}x_1x_2 + x_2^2\end{array}$
					\end{center}
		\item We simply need to substitute $x_1=6$ and $x_2=1$ to obtain $Q(6,1)=185$.
		\item $Q(1,3)=16$.
	\end{enumerate}
}
\useproblem{lay:7_2_1}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

